Objective NCERT GEAR UPCHEMISTRY

Altis Vortex New Delhi

Some Basic Concepts of Chemistry 1

Structure of Atom 27Classification of Elements and Periodicity in

Properties 56

Chemical Bonding and Molecular Structure 76

States of Matter 99

Thermodynamics 124

Equilibrium 151

Redox Reactions 184

Hydrogen 203

The s-Block Elements 217

The p-Block Elements 234

Organic Chemistry- Some Basic Principles and Techniques 251

Hydrocarbons 278

Environmental Chemistry 304

The Solid State 1

Solutions 33

Electrochemistry 64

Chemical Kinetics 90

Surface Chemistry 120

General Principles and Processes of Isolation of Elements 142

The p-Block Elements 158

The d-and f-Block Elements 179

Coordination Compounds 196

Haloalkanes and Haloarenes 215

Alcohols, Phenols and Ethers 247

Aldehydes, Ketones and Carboxylic Acids 279

Amines 310

Biomolecules 339

Polymers 357

Chemistry in Everyday Life 373

Class XI

Class XII

Contents

AIIMS is every medical aspirants dream and true to its name, Aim4AIIMS provides exact guidance on how to aim for and achieve that dream.I had been hearing the Toppers interviews since 2014 and that gave me great direction on how to prepare, what is AIIMS pattern like and the importance of NCERT text books.Joining the Aim4AIIMS classes in October 2015 turned out to be the best decision of my life. I got interact with previous years toppers where they guided us on their method of preparation and how to approach the exam. I am deeply grateful to Dr. Ajay Mohan Sir and the entire team of Aim4AIIMS for their guidance and support, due to which i was able to achieve my dream of AIIMS-New Delhi.Thank you Aim4AIIMS for helping me aim for, and hit the target!!

IRA PACHORIAIR-11, AIIMS-2016

AIR-43 JIPMER-2016

I am really happy that all my hard work has paid off. I would thank my parents for this success as it would not be possible without their support.The Aim4AIIMS initiative proved to be boon to me. It gave me a clear idea of the kind of questions asked in exam. Vashishta Sir and Ansh sir helped me a lot in my preparation. Dr. Ajay Mohan sir's guide to AIIMS preparation really helped me out in the GK section as many as 15 questions could be answered with the information on current affairs in the book and posted on the website.I wish all the AIIMS aspirants good luck for their exam.

SATHVIK REDDY ERLAAIR-1, AIIMS-2016

AIR-20, JIPMER-2016

I am so happy that I secured 9th Rank in AIIMS and 5th Rank in NEET. It all happened because I studied to the NCERT and prefer supplement which elaborates the NCERT. So, i will suggest everyone study NCERT. This year I found that 90% to 95% MCQs in biology had been asked in the Exam. So, for Biology NCERT is most important. You need to pay more attention to the language and examples of NCERT because many questions come from these examples.Many aspirants think that NCERT is only important for biology. But it is not true. It is also important for chemistry and physics. NCERT is really important for Surface and state of matter in Physical chemistry.For inorganic chemistry, you need to read every reaction, example and experimental data given in NCERT. So read and analyze line by line and revise repeatedly.Coming to the organic chemistry which was really hard for me personally, for some topics like Environmental Chemistry, Biomolecules, Polymers and Chemistry in Everyday Life, you really need to read NCERT. Solve examples given in the NCERT which would be really helpful for the exam.In Physics, NCERT is very important for A & R questions. Analyze logically to the concepts and solved examples given in the NCERT.At the last, once again I would like to say that “Never ignore NCERT if you really want to secure good Rank”.

ABHISHEK DOGRAAIR-9, AIIMS-2017

AIR-5, NEET-2017

TESTIMONIES

2 Structure of Atom

J.J. Thomson proposed the first atomic model known as the Plum Pudding model.

Rutherford's Atomic ModelRutherford proposed the nuclear model of atom. According to this model,

• The positive charge and most of the mass of the atom was densily concentrated in extremely small region called nucleus.

• The nucleus is surrounded by electrons that move around the nucleus in circular path called orbits.

• Electrons and nucleus are held together by electrostatic force of attraction.

Atomic Number and Mass Number• Atomic Number: The number of protons present in the

nucleus of an atom is known as Atomic Number. It is represented by "Z".

• Mass Number: The mass of the nucleus is due to the presence of protons and neutrons, thus, they are called as nucleons. It is represented by "A".

Photoelectric Effect

The Photoelectric effect is the observation that many metals emit electrons when light is focussed upon them.

20

0

1E = W mv2where; E = Energy of photon W Work function

+

=

Bohr's ModelBohr gave the model of stationary energy orbits within the atom.

He also gave emission spectrum of hydrogen atom.

2H 2 2

1 2

7 1H

1 1 1=

R 1.097 10 −

ν = − λ

= ×

R Z sn n

m

• For Lyman series; n1 = 1 ; n2 >1

• For Balmer series; n1 = 2 ; n2 > 2

• For Paschen series; n1 = 3 ; n2 > 3

• For Brackett series; n1 = 4 ; n2 > 4

• For Pfund series; n1 = 5 ; n2 > 5

• For Humphery series; n1 = 6 ; n2 > 6

Limitations of Bohr's Model• Zeeman Effect: Splitting of spectral line by magnetic

field.

• Stark Effect: Splitting of spectral line by electric field.

De-Broglie RelationshipMatter which exhibit both particle as well as wave nature is called as de-broglie relationship.

h hp mv

λ = =

Heisenberg's Uncertainty PrincipleAccording to this principle, it is impossible to determine simultaneously the exact position and exact momentum (or velocity) of an electron.

Quantum Mechanical ModelA model plane or surface is the plane in which the probability of finding an electron is zero.

Critical Points

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The energy required to remove the electron from the outermost shell is called ionisation energy.

Pauli's Exclusive PrincipleIt states that "no two electrons in the same atom can have the same set of quantum number."

Hund's RuleAccording to this rule of maximum multiplicity, electrons enter one by one in the orbitals of a subshell remain unpaird as far as possible.

Aufbau PrincipleAccording to this principle, electron enters in the subshells of an atom in the increasing order of energy.

Quantum Number• Principle Quantum Number (n): Shows energy level

• Angular/Azimuthal Quantum Number(l): Show subshell (orbital)l = 0 = s - orbital l = 1= p - orbitall = 2 = d - orbital l = 3 = f - orbital

• Magnetic Quantum Number (m): Shows subshell (orbital) preferred orientation in 3-D shape.

• Spin Quantum Number (s): Shows the direction of spin of electrons.

1s = 2±

Critical Points of Atomic Structure

• Moseley postulated that the frequency of the X-rays was related to the charge present on the nucleus of an atom of the element used as anticathode and found that

a(Z-b)frequency

Z= nuclear chargea,b = constant

ν =ν = a = proportionality constant

b = Screening constant

• Yukawa(1935) suggested that a pair of nucleus is held together by continuously exchanging their charge through the agency of particles called mesons (π) which may be electrically neutral, positive or negative.

• Mesons are unstable particles about 200 times as heavy as an electron and it can be found in cosmic rays.

• Cathode rays produces X-rays when they strike a metallic target.

• J.C. Maxwell in 1864, suggested than an alternating current of high frequency is capable of radiating energy in the form of waves which travel in space with a same speed.

• Louis de-broglie's concept of wave nature of electron was experimentally verified by Davisson and Germer in 1927.

Ritz Combination Principle

States that "the wave number of any line in the hydrogen spectrum of a particular series can be represented as a difference of the two terms i.e., one is a constant and the other varies throughout the series.

Mathematically ; 2 21 1Rx y

− ν = −

R = Rydberg Constantx & y = Integersy is always greater than x

• The atomic spectrum of hydrogen is simplest of line spectra.

• Bohr's model of circular orbit was extended by Sommerfeld by introducing the concept of elliptical orbits. The principal quantum number (n), is used by Bohr and azimuthal quantum number (κ) used by Sommerfeld are related to

one another as; n length of major axis of elliptical orbitlength of minor axis of elliptical orbit

=κ

Compton Effect

Compton effect is when monochromatic X-rays are allowed to fall on some light elements i.e., X-rays interact with the electrons, the scattered X-rays have longer wavelength or less frequency or less energy than the incident rays.

Probability Distribution Curve:

The probability of locating the electron at different distances from the nucleus can be plotted graphically probability (ψ2) against the distance (r) from the nucleus of the atom.

• The plane on which the probability of finding the electron is zero is known as nodal plane.

• The distance of maximum probability for 1s e– of hydrogen atom is 0.59 Å and it is equal to Bohr's radius for the first orbit.

• Spectroscopic studies, which are used in elucidating electronic configuration, show that four numbers known as quantum number are required to characterize each electron in an atom.

• The passage of electricity through a gas, called a discharge, was familiar to Michael Faraday, but steady conduction takes places when the pressure of the gas is less than about 50 mm Hg.

• One faraday is the charge on one mole (Avogadro's number) of an electrons which is 96485 coulomb.

• Modern medical practice would not be done without X-ray images. In addition to their medical value, X-rays

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Structure of Atom

are also extremely useful in the determination of the structure of molecules.

• The Daltonian idea for the individuability of atoms is a contrary of radioactivity.

• An element whose all atoms have the same atomic number is a pure substance.

• All three quantities, i.e., atomic number, mass number and neutron number must be positive integers or whole numbers.

• Because the mass of the proton and the neutron are both about 1 atom, whereas the mass number is appoximately equal to actual mass of isotope.

• Light wave is an oscillating electric and magnetic fields that can travel through space.

• Microwave ovens represent an interesting domestic application of these waves.

• Eye's senstivity is a function of wavelength, the sensitivity being the maximum at a wavelength of about 5.6×10-7 m(yellow-green).

• Nanometers is the unit for measure of the wavelength of an electromagnetic radiations.

• Radiation emitted from the human body and warm objects is mostly infrared, which is detected by burglar alarms, military night vision scoper, and similar equipment.

• One electron volt (eV) is the energy that a material particle with same charge as an electron (or a proton) picks up when it falls through a potential drop of one volt. Recall that 1 coulomb × 1 volt = 1 Joule

Then; l eV = (1.6 × 10–19C)(1V) = 1.602 × 10–19J

• Continous spectra is given by incandescent ("red hot" or "white hot") solids, lights and high pressure gases.

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Sub-atomic particles [NCERT Pg 27-29]

1. The charge of an electron was discovered by:a. J.J. Thomson b. Neil Bohrc. Chadwick d. Millikan

2. Particle nature of electron was discovered by:a. Maxborn b. J.J. Thomsonc. De-Broglie d. Schrodinger

Atomic Models [NCERT Pg 29-34]

3. Be2+ is isoelectronic with which of the following ions:a. H+ b. Li+ c. Na+ d. Mg2+

4. (32Ge76, 34Se

76) and (14Si30, 16S

32) are the examples of-a. Isotopes and isobars b. Isobars and isotonesc. Isotones and isotopes d. Isobars and isotopes

5. The lightest particle is:a. α-particle b. Positronc. Proton d. Neutron

6. The ratio of charge and mass would be greater for:a. Proton b. Electronc. Neutron d. α-particle

Developments Leading to the Bohr's Model of Atom [NCERT Pg 34-42]

Formulas

• Photo-electric effect: 20

1E = W mv2+

• Planck's equation: hcE = h = K Eν ∴ ⋅ ⋅ ∝ νλ

• Frequency: H 2 21 2

RE 1 1c = h ,h h

∆ν ν = = −

n n

7. The scientist who proposed the atomic model based on the quantisation of energy for the first time is:a. Max Planck b. Neil Bohrc. De-Broglie c. Heisenberg

8. The Wavelength (in Å) of an emission line obtained for Li2+ during a electronic transition from n2 = 2 to n1 = 1 is: (R = Rydberg constant)a. 3R/4 b. 27R/4c. 4/3R d. 4/27R

9. Splitting of spectral lines under the influence of magnetic field is called:a. Zeeman effect b. Stark effectc. Photoelectric effect c. None of these

10. An element X has the following composition 200X:90%, 199X:8.0%, 202X:2.0%. The average atomic mass of the naturally occurring element X is closest to:a. 200 u b. 201 uc. 202 u d. 199 u

11. The value of Rydberg constant is:a. 109, 677 cm–1 b. 19876 cm–1c. 108769 cm–1 d. 108976 cm–1

12. Which pair is of isoelectronic species?a. K+, Cl–, Na+ b. K +, Cl–, Ca2+c. Fe–, Al3+, K+ d. Fe2+, Cu2+, V3+

13. What is the lowest energy of the spectral line emitted by the hydrogen atom in the Lyman series?

a. H5hR c36

b. H4hR c3

c. H3hR c4

d. 7Rh144

Bohr's Model for Hydrogen Atom [NCERT Pg 42-46]

Formulas

• Angular Momentum: nhmvr =2π

n = 1, 2, 3.....

• Bohr's Radius: 2

0n

2 0

r nr

n0.529 A

Z

Z

=

= ´

• Energy of Electron in nth Bohr's orbit: 2

n 2

13.6 eVEn

Zatom

-=

• Potential Energy: 2

P0

Ze(E ) =4 r−π ε

• Speed of electron in nth orbit (Vn): 6

n2.18 10 ZV m/s

n´

=

Self Assessment Questions

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Structure of Atom

• Number of revolutions made by an electron in nth Bohr's

orbit: nV round/ sec2 r

=π

• Centrifugal force of orbitng electrons = Coulombic

attraction by nucleus: 2 2

20

mv zei.e, 4 E r

=πr

• Wave Number: 2H 2 21 2

7 1H

1 1 1R Zn n

R 1.0967 10 m

−

−

ν = = − λ

= ×

• Energy of electron in its stationary state

n H 2

1E Rn

n 1,2,3.....

= −

=

14. The emission spectrum of hydrogen discovered first and the region of the electromagnetic spectrum to which it belongs, respectively are-a. Lyman, Ultravioletb. Lyman, Visiblec. Balmer, Ultravoilet d. Balmer, Visible

15. The velocity of electron in second shell of hydrogen atom is:a. 10.94 × 106 m/sec b. 18.88 × 106 m/secc. 1.88 × 106 m/sec d. 1.094 × 106 m/sec

16. For which of the following species, Bohr's theory is not applicable?

a. Be3+ b. Li2+

c. He2+ d. H

17. If the first ionisation energy of H- atom is 13.6 eV, then the second ionisation energy of He- atom is:a. 27.2 eV b. 40.8 eVc. 54.4 eV d. 108.8 eV

18. When the electrons of hydrogen atom return to L-shell from shell of higher energy, we get a series of lines in the spectrum. This series is called:a. Balmer series b. Lyman seriesc. Brackett series c. Pascher series

19. The nitride ion in lithium nitride is composed of:a. 7-protons + 10 electrons b. 10-protons + 10 electronsc. 7-protons + 7 electrons d. 10-protons + 7 electrons

20. In which one of the following pairs, the two species are both isoelectronic and isotopic? (Atomic number: Cr = 24, Ar = 18, K = 19, Mg = 12, Fe = 26, Na = 11)a. 24Mg2+ and 40Ar b.

39K+ and 40K+

c.24Mg2+ and 25Mg d. 23Na and 24Na+

21. The ratio of neutrons in C and Si with respective atomic masses 12 and 28 is:a. 2:3 b. 3:2c. 3:7 d. 7:3

22. If a species has 16 protons, 18 electrons and 16 neutrons. Find the species and its charge:a. S– b. Si2– c. S3– d. S2–

23. When the electron of a hydrogen atom jump from n = 4 to n = 1 state, the number of different spectral line emitted are:a. 15 b. 9 c. 6 d. 3

24. The wave number of the spectral line in the emission spectrum of hydrogen will be equal to 8/9 times the rydberg constant if the electron jumps from:a. n = 3 to n = 1 b. n = 10 to n = 1c. n = 9 to n = 1 d. n = 2 to n = 1

25. The radius of the first Bohr orbit of the H-atom is r. Then, the radius of the first orbit of Li2+ will be:a. r/9 b. r/3 c. 3r d. 9r

26. In which one of the following, the number of protons is greater than the number of neutrons but number of protons is less than the number of electrons?a. D3O

+ b. SO2d. H2O d. OH

–

27. The energy ratio of a photon of wavelength 3000Å and 6000Å is:a. 1:1 b. 2:1 c. 1:2 d. 1:4

28. Bohr's model can explain:a. The solar spectrumb. Spectrum of hydrogen moleculec. Spectrum of any atom or ion containing one electron

onlyd. Spectrum of hydrogen atom only

29. The first line emission of hydrogen atom spectrum in the Balmer species appears at:a. 5R/36 cm–1 b. 3R/4 cm–1c. 7R/144 cm–1 d. 9R/400 cm–1

30. The radius of 2nd Bohr's orbit of hydrogen atom is:a. 0.053 nm b. 0.106 nmc. 0.2116 nm d. 0.4256 nm

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31. The maximum energy possessed by an electron in when it is present:a. In nucleusb. In ground energy statec. In 1st excited stated. At infinite distance from the nucleus

32. Which one of the following transition have minimum wavelength?a. n4 → n1 b. n2 → n1c. n4 → n2 d. n3 → n1

33. Select the pair where both species have same radius?a. r4He

+ and r1H b. r2Be3+ and r1H

c. r2He and r1H d. r3Li2+ and r1H

34. The radius of the first orbit of hydrogen atom is 0.52 × 10–8 cm. The radius of the first orbit of He+ ion is a. 0.26 × 10–8 cm b. 0.52 × 10–8 cm c. 1.04 × 10–8 cm d. 2.08 × 10–8 cm

35. The ratio of ionization energy of H and Be+3 is a. 1 : 1 b. 1 : 3 c. 1 : 9 d. 1 : 16

36. As we move away from nucleus, the energy of orbit a. Decreases b. Increases c. Remains unchanged d. None of these

37. In hydrogen atom, energy of first excited state is –3.4 eV. Then find out KE of same orbit of hydrogen atoma. +3.4 eV b. +6.8 eV c. –13.6 eV d. +13.6 eV

Towards Quantum Mechanical Model of the Atom [NCERT Pg 46-49]

Formulas

• Heisenberg's Uncertainty Principle: hp4

∆ ×∆ ≥π

x

• Number of Spectral Line: 2 1 2 1(n n )(n n 1)2

- - -

• De-Broglie Relationship: h h hp mv 2mE

λ = = =

38. Which among the following species have the same number of electron in its outermost as well as penultimate shell?a. Mg2+ b. O2–c. F– d. Ca2+

39. The uncertainty in the velocity of particle of mass 6.626×10–28 Kg is 106 m/sec. What is the uncertainty in its position in nm?

a. 12π

b. 2.5π

c. 4π

d. 14π

40. The ratio of de-broglie wavelength of a deuterium atom to that of α-particle when the velocity of former is five times greater then that of latter is:a. 4 b. 10.2 c. 2 d. 0.4

41. As per de-broglie formula a microscopic particle of mass 100 g and moving at a velocity of 100 cm/sec will have a wavelength of:a. 6.6 × 10–29 cm/s b. 6.6 × 10–30 cm/sc. 6.6 × 10–31 cm/s d. 6.6 × 10–32 cm/s

42. The de-broglie wavelength associated with particle of mass 10–6 kg moving with a velocity of 10 m/sec is:a. 6.63 × 10–7 m b. 6.63 × 10–16 mc. 6.63 × 10–21 m d. 6.63 × 10–29 m

43. If Ee, Eα, and Ep represents the kinetic energy of an electron, α-particle and proton respectively each moving with same de-broglie wavelength then:a. Ee = Eα = Ep b. Ee > Eα > Epc. Eα > Ep > Eα d. Ee > Ep > Eα

44. In sommerfeld's modification of Bohr's theory the trajectory of an electron of hydrogen atom is:a. A perfect ecllipseb. A close ecllipse like curve, narrower at the perihelion

position and flatter at the aphelion positionc. A close loop on spherical surfaced. Rosette

45. Uncertainty principal was given by-a. Einstein b. Heisenbergc. Rutherford d. Thomson

46. A particle having a mass of 1 mg has a velocity 1 Km/sec. Calculate the wavelength of the particle:a. 6.626 × 10–28 cm b. 6.626 × 10–29 cmc. 6.626 × 10–30 cm d. 6.626 × 10–31 cm

47. The uncertainty in the position of an electron moving with a velocity of 3 × 104 cm/sec accurate upto 0.011% will be:a. 1.92 cm b. 7.66 cmc. 0.175 cm d. 3.84 cm

48. If uncertainty in position and velocity are equal, then uncertainty in momentum will be:

a. 1 mh2 π

b. 1 h2 mπ

c. h4 mπ

d. mh4π

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Structure of Atom

49. The velocity of particle "A" is 0.1m/sec and that of particle "B" is 0.05m/sec. If the mass of particle B is 5 times that of particle A, then the ratio of de-broglie wavelength associated with particle A and B:a. 2:5 b. 3:4 c. 6:4 d. 5:2

50. If de-broglie wavelength of mass 'm' is 100 times of its velocity then its value in term of its mass 'm' and planck's constant "h" is:

a. 1 m10 h

´ b. h10m

c. 1 h10 m

d. m10h

51. The wavelength of electron waves in two orbit is in ratio 3:5. The ratio of K.E. of electrons will be:a. 25:9 b. 15:3 c. 9:25 d. 3:5

52. If uncertainty in position and momentum are equal, then uncertainly in velocity is:

a. 1 h2m π

b. h2π

c. 1 hm π

d. hπ

53. The de-broglie wavelength associated with a mass of 1kg having K.E. 0.5J is:a. 6.626 × 10–34 m b. 13.20 × 10–34 mc. 10.38 × 10–31 m d. 6.626 × 10–34 Å

54. Wave nature of electron was demonstrated by:a. Schrondinger b. De-Brogliec. Davison and Germer d. Heisenberg

55. For an electron, if the uncertainty in velocity is ∆v, the uncertainty in position ∆x is given by:

a. hm4 vπ∆

b. 4hm v

π∆

c. h4 m vπ ∆

d. 4 mh vπ∆

56. Which have the same number of s-electrons as the d-electrons in Fe2+?a. Li b. Na c. N d. P

57. The number of waves made by an electron moving in an orbit having maximum magnetic quantum number + 3 is -a. 4 b. 3 c. 5 d. 6

58. The set of quantum no. not applicable for an electron -

a. 12,0,0,2

− b. 13,1, 2,2

− +

c. 14,2, 2,2

− − d. 16,0,0,2

+

Quantum Mechanical Model of Atom [NCERT Pg 49-60]

FormulasTotal number of nth orbital = n2

Total number of electron in nth orbital = 2n2

Azimuthal quantum number (): = n – 1• Magnetic Quantum Number:

Total number of ml = 2l + 1Total number of nodes = n – 1Total number of radial nodes = n – l – 1

• Orbital angular momentum (l): h( +1)2π

• Magnetic Moment n(n+2) BMB( ) :µ

59. The quantum of light energy is called:a. Photon b. Neutronc. Electron d. Proton

60. Magnitude of kinetic energy in an orbit is equal to:a. Half of the P.E.b. Twice of the P.E.c. 1/4th of the P.E.d. None of the above

61. What is the orbital angular momentum of an electron in f-orbital?

a. 1.5hπ

b. 6hπ

c. 3hπ

d. 3hπ

62. The magnetic moment of Ni2+ ion in BM unit is:a. 1.73 b. 4.81 c. 5.96 d. 2.83

63. The maximum number of electrons which can be held by subshell with azimuthal quantum number "l" in an atom is given by: a. 2 + 1 b. 2 + 2c. 2(2 + 1) d. 2(2 + 2)

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64. What is the maximum number of electrons that can be associated with a following set of quantum number n = 3, l = 1 and m = –1?a. 10 b. 6 c. 4 d. 2

65. The quantum number "m" of a free gaseous atom is associated with:a. Effect volume of orbitalb. Shape of orbitalc. Spatial orientation of orbitald. Energy of orbital in the absence of magnetic field

66. Which of the following element is represented by electronic configuration 2 2 1 1 1x y z1s 2s 2p 2p 2p ?a. N b. O c. P d. S

67. The atomic number of element is 17. The number of orbital containing electron pair in its valence shell is:a. 3 b. 4 c. 6 d. 8

68. The total number of electrons present in all the p– orbital of bromine are-a. 5 b. 15 c. 17 d. 35

69. The number of unpaired electron in Fe3+(Z = 26) are:a. 5 b. 6 c. 3 d. 4

70. The orbital angular momentum of p- electron is given as:

a. 2π b. h2π

c. 32π

h d. πh

71. Which of the following statement in relation to the hydrogen atom is correct:a. 3s, 3p and 3d orbital all have the same energyb. 3s and 3p orbital are of lower energy than 3d orbitalc. Both of thesed. None of these

72. According to Bohr's theory, the angular momentum of an electron in the 4th orbit is:

a. h2π

b. 2hπ

c. 3h2π

d. 3hπ

73. The total number of orbitals in a shell with principle quantum number n is:a. 2n b. 2n2c. n2 d. n + 1

74. If the radius of 1st Bohr orbit be a0, then radius of 3rd Bohr

orbit would be:a. 3a0 b. 6a0

c. 9a0 d. 19 0

a

75. Which of the following is not possible?a. n = 3, = 0, m = 0 b. n = 3, = 1, m = –1c. n = 2, = 0, m = –1 d. n = 2, = 1, m = –1

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Structure of Atom

1. If the binding energy of 2nd excited state of H like sample is 24 eV approximately, then the ionisation energy is approximately:a. 54.4 eV b. 24 eVc. 122.4 eV d. 216 eV

2. The de-broglie wavelength of an electron moving in a circular orbit is λ.

The minimum radius of the orbit is given by:

a. λπ

b. 2λπ

c. 4λπ

d. 3λπ

3. A sodium street light gives off yellow light that has a wavelength of 600 nm. Then:

0

0

12400eV AEnergyof photon(A)

= λ

a. Frequency of this light is 7 × 1014 s–1b. Frequency of this light is 5 × 1014 s–1c. Energy of photon is approximately 2.07 eVd. Both (b) and (c)

4. Photon having wavelength 310 nm is used to break the bond of A2 molecule having bond energy 288 kJ/mol, then the percentage of energy of photon converted to KE is (1 eV = 96 kJ/mol):a. 25 b. 50 c. 75 d. 80

5. The radii of two of first six Bohr’s orbits of H atom are in the ratio 4 : 9. The energy difference between them may be:a. Either 12.09 eV or 3.4 eVb. Either 2.55 eV or 10.2 eVc. Either 1.89 eV or 0.49 eVd. 1.89 eV only

6. In which transition, one quantum of energy is emitted?a. n2 = 4 → n1 = 2b. n2 = 3 → n1 = 1c. n2 = 2 → n1 = 1d. All of these

7. Last line of Lyman series for H-atom has wavelength λ1 Å. The 2

nd line of Balmer series has wavelength λ2 Å, then:

a.1 2

16 9=

λ λ b.

2 1

16 3=

λ λ

c.1 2

4 1=

λ λ d.

1 2

16 3=

λ λ

8. Which of the following represents the correct set of quantum numbers of a 4d electron?a. 4, 3, 2, +1/2 b. 4, 2, 1, 0c. 4, 3, –2, +1/2 d. 4, 2, 1, –1/2

9. Which of the following statements is/are correct for an electron having quantum numbers n = 4 & m = 2?a. The value of may be 2b. The value of may be 3c. The value of s may be +1/2d. All of these

10. Change in orbit angular momentum when an electron makes a transition corresponding to 3rd line of Balmer series in Li2+ ion is:

a. h2π

b. 2h2π

c. 3h2π

d. 4h2π

11. An electron in a hydrogen like atom makes transition from a state in which its de-brogile wavelength is λ1 to a state where its de-brogile wavelength is λ2 then wavelength of photon (λ) generated will be:a. λ = λ1 – λ2

b. 2 2

1 22 2

1 2

4mch

λ λλ = λ − λ

c. 2 2

1 22 2

1 2

λ λλ =

λ −λ

d. 2 2

1 22 2

1 2

2mch

λ λλ = λ − λ

12. The ionisation energy of H atom is x J/atom. The wavelength of first Balmer line for He+ ion is-

a. 5x36 b.

5x9

c. 5x

36hc d.

9hc5x

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13. An energy of 24.6 eV is required to remove one of the electrons from a He atom. The energy required to remove both the electrons from He atom is-a. 38.2 eV b. 49. 2 eVc. 51.8 eV d. 79 eV

14. Photons of frequency 3.2 × 1016 Hz is used to irradiate a metal surface, the maximum kinetic energy of the emitted photo-electron is

th34

of the energy of the

irradiating photon? What is the threshold frequency of

the metal?a. 2.4 × 1025 Hz b. 2.4 × 1016 Hzc. 1.6 × 1015 Hz d. 8 × 1015 Hz

15. The total number of orbitals in the principal shell of He+

that has energy equal to Rhc4

− (where R is Rydberg’s constant).

a. 4 b. 8c. 16 d. 32

16. Not considering the electron spin, the degeneracy of second excited state is 9, while the degeneracy of first excited state of H– is-a. 1 b. 2 c. 3 d. 4

17. Not considering the electron spin, the degeneracy of the first excited state (n = 2) of H atom is-a. 2 b. 3 c. 4 d. 8

18. If ΔE is the energy emitted in eV when an electronic transition occurs from higher energy level to lower energy level in H–atom, the λ of the line produced is approximately equal to-

a. o12375 A

E∆ b.

o19800 AE∆

c. o13600 A

E∆ d.

o21800 AE∆

19. Which of the following is the energy of a possible excited state of hydrogena. +13.6 eV b. –3.4 eVc. +6.8 eV d. –2.8 eV

20. If an electron travels with a velocity of 1 th

100 speed of

light in Bohr’s first orbit, what is its velocity (relative to the speed of light) in the 5th Bohr orbit?a. 0.002 b. 0.1c. 0.5 d. 0.7

21. The ionisation energy of He+ is 19.6 × 10–18 J atom–1. The energy of the first stationary of Li2+ is-a. –2.2 × 10–15 J atom–1b. 8.82 × 10–17 J atom–1c. 4.41 × 10–16 J atom–1d. –4.41 × 10–17 J atom–1

22. Photoelectric emission is observed from a metal surface with incident frequencies ν1 & ν2 where ν1> ν2. If the kinetic energies of the photoelectrons emitted in the two cases are in the ratio 2:1, then the threshold frequency νo of the metal is-a. 2ν2 – ν1 b. 2ν1 – ν2

c. 1 22

ν − ν d. ν2 – ν1

23. The ratio of area covered in 2nd orbit to first orbit is-a. 1 : 1 b. 16 : 1 c. 1 : 16 d. 4 : 1

24. If ro be the radius of first Bohr's orbit of H-atom, the de-Broglie's wavelength of an electron revolving in the third Bohr's orbit will be:a. 2πro b. 4πro c. 6 πro d. πro

25. An electron in an atom undergoes transition in such a way that its kinetic energy changes from x to x ,

4the

change in potential energy will be :

a. 3 x2

+ b. 3 x8

−

c. 3 x4

+ d. 3 x4

−

26. The wavelength of first line of Balmer spectrum of hydrogen will be :a. 4340 Å b. 4101 Å c. 6569 Å d. 4861 Å

27. What transition in He+ ion shall have the same wave number as the first line in Balmer series of hydrogen atom ?a. 3 → 2 b. 6 → 4 c. 5 → 3 d. 7 → 5

28. The wavelength of electron waves in two orbits is 3 : 5. The ratio of kinetic energy of electron will be a. 25 : 9 b. 5 : 3 c. 9 : 25 d. 3 : 5

29. In an atom, an electron is moving with a speed of 600 m/s with an accuracy upto 0.005%. What is the uncertainty in position?a. 1.52 × 10–4 m b. 5.10 ×10–3 mc. 1.92 × 10–3 m d. 3.84 × 10–3 m

30. The energy absorbed by each molecule A2 of a substance is 4.4 × 10–19 J and bond energy per molecule is 4 × 10-19 J. the K.E. of the molecule per atom is-a. 2.2 × 10–19J b. 2 ×10–19Jc. 4 × 10–20J d. 2 × 10–20J

31. What is the maximum number of electrons which can be accomodated in principal quantum number 4?a. 16 b. 18 c. 22 d. 32

32. What is the maximum number of electrons which can be accomodated in an atom in which the highest principal quantum no. is 4?a. 10 b. 18 c. 36 d. 54

37

Structure of Atom

33. How many elements would be in the IInd period of the Periodic Table if the spin quantum numbers could have the

value 1 1, 0,2 2

+ − ?

a. 18 b. 12 c. 10 d. 8 34. Suppose that a hypothetical H-like atom gives a red, green,

blue and violet line in spectrum. Which jump, according to figure, would give off the violet spectral line;

a. 3 → 1 b. 2 → 1 c. 4 → 1 d. 3 → 2 35. Which of the following set of quantum number is

possible?a. n = 4, l = 2, m = –2, s = –2 b. 1n 4, l 4, m 0,s

2= = = =

c. 1n 4, l 3, m 3,s2

= = = − =

d. n = 4, l = 0, m = 0, s = 0 36. In an atom, which has 2K, 8L, 18M and 2N electrons in

the ground state. The total number of electrons having magnetic quantum number, m = 0 isa. 6 b. 10 c. 7 d. 14

37. The probability density curve for 2s electron appears like

a. b.

c. d.

38. Which of the following statement concerning the four quantum number is incorrect?a. n gives the size of an orbitalb. l gives the shape of an orbitalc. m gives the energy of the electron in orbitald. Both (a) & (b)

39. Consider the following sets of quantum numbern l m s

(i) 3 0 0 +1/2(ii) 2 2 1 +1/2(iii) 4 3 –2 –1/2(iv) 1 0 –1 –1/2(v) 3 2 3 +1/2

Which of the following sets of quantum number is not possible?a. (i), (ii), (iii) and (iv) b. (ii), (iv) and (v) c. (i) and (iii) d. (ii), (iii) and (iv)

40. Any f-orbital can accommodate uptoa. 2 electrons with parallel spin b. 6 electronsc. 2 electrons with opposite spin d. 14 electrons

41. Which of the following sets of quantum number is/are incorrect ?a. 1n 3, 3,m 0, s

2= = = =

b. 1n 3, 2,m 2, s2

= = = = −

c. 1n 3, 1,m 2, s2

= = = = −

d. 1n 3, 0,m 0, s2

= = = = +

42. If an electron in H atom has an energy of –78.4 kcal/mol. The orbit in which the electron is present is :-a. 1st b. 2nd c. 3rd d. 4th

43. If the radius of 2nd Bohr orbit of hydrogen atom is r2. The radius of third Bohr orbit will be :-a. 2

4 r9

b. 4r2 c. 29 r4

d. 9r2

44. Difference between nth and (n + 1)th Bohr's radius of H–atom is equal to its (n – 1)th Bohr's radius. The value of n is :-a. 1 b. 2 c. 3 d. 4

45. The dissociation energy of H2 is 430.53 kJ mol–1. If H2 is

dissociated by illumination with radiation of wavelength 253.7 nm. The fraction of the radiant energy which will be converted into kinetic energy is given by :-a. 8.86% b. 2.33% c. 1.3% d. 90%

46. Light of wavelength λ shines on a metal surface with intensity x and the metal emits Y electrons per second of average energy, Z. What will happen to Y and Z if x is doubled ?a. Y will be double and Z will become half b. Y will remain same and Z will be doubledc. Both Y and Z will be doubled d. Y will be doubled but Z will remain same

47. Select the wrong statement (s) from the following?a. If the value of = 0, the electron distribution is sphericalb. The shape of the orbital is given by magnetic quantum

number

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c. Angular momentum of 1s, 2s, 3s electrons are equald. (b) and (c) both are correct

48. Which one is the wrong statement?a. The uncertainty principle is hE t

4∆ ×∆ ≥

πb. Half filled and fully filled orbitals have greater stability

due to greater exchange energy, greater symmetry and more balanced arrangement.

c. The energy of 2s-orbital is less than the energy of 2p-orbital in case of hydrogen like atoms.

d. de-Broglie’s wavelength is given by hm

λ =ν

, where

m = mass of the particle, v = group velocity of the particle.

49. An isotone of 7632 Ge is i. 7732 Ge ii.

7733 As

iii. 7734 Se iv. 7834Sea. Only (i) and (ii) b. Only (ii) and (iii)c. Only (ii) and (iv) d. Only (ii), (iii) and (iv)

50. Arrange the electrons represented by the following set of quantum numbers in the decreasing order of energy(i) n = 4, l = 0, m = 0, s = + 1/2(ii) n = 3, l = 1, m = 1, s = – 1/2(iii) n = 3, l = 2, m = 0, s = + 1/2(iv) n = 3, l = 0, m = 0, s = – 1/2(a) (i) > (ii) > (iii) > (iv)(b) (iv) > (iii) > (ii) > (i)(c) (iii) > (i) > (ii) > (iv)(d) (i) > (iii) > (ii) > (iv)

51. Select the correct statement(s):a. The value of spin only magnetic moment of Co3+ ion

(in BM) = 24b. The number of radial nodes in a 3p-orbital = 1c. The number of electrons with (m = 0) in Mn2+ ion = 11d. All are correct

Image Based Questions

52. The plot of orbital have function ψ(r) as a function of distance r of the electron from the nucleus for 2s orbital is.

a.

b.

c.

d.

53. The diagram shows the energy levels for an electron in certain atom. Which transition shown represents the emission of a photon with the most energy?

a. II b. I c. IV d. III

54. Which of the following radial distribution graphs corresponds to l = 2 for H atom for the least value of n for which l = 2 is allowed?

a. b.

c. d.

39

Structure of Atom

1. Which of the following conclusions could not be derived from Rutherford’s α-particle scattering experiment?a. Most of the space in the atom is emptyb. The radius of the atom is about 10–10 m while that of

nucleus is 10–15 mc. Electrons move in a circular path of fixed energy

called orbitsd. Electrons and the nucleus are held together by

electrostatic forces of attraction

2. Which of the following options does not represent ground state electronic configuration of an atom?

a. 1s2 2s2 2p6 3s2 3p6 3d8 4s2

b. 1s2 2s2 2p6 3s2 3p6 3d9 4s2

c. 1s2 2s2 2p6 3s2 3p6 3d10 4s1

d. 1s2 2s2 2p6 3s2 3p6 3d5 4s1

3. The probability density plots of 1s and 2s orbitals are given in figure:

The density of dots in a region represents the probability density of finding electrons in the region.

On the basis of above diagram which of the following statements is incorrect?a. 1s and 2s orbitals are spherical in shapeb. The probability of finding the electron is maximum

near the nucleusc. The probability of finding the electron at a given

distance is equal in all directionsd. The probability density of electrons for 2s orbital

decreases uniformly as distance from the nucleus increases

4. Which of the following statement is not correct about the characteristics of cathode rays?

a. They start from the cathode and move towards the anode

b. They travel in straight line in the absence of an external electrical or magnetic field

c. Characteristics of cathode rays do not depend upon the material of electrodes cathode ray tube

d. Characteristics of cathode rays depend upon the nature of gas present in the cathode ray tube

5. Which of the following statements about the electron is incorrect?a. It is a negatively charged particleb. The mass of electron is equal to the mass of neutronc. It is a basic constituent of all atomsd. It is a constituent of cathode rays

6. Which of the following properties of atom could be explained correctly by Thomson model of atom?a. Overall neutrality of atomb. Spectra of hydrogen atomc. Position of electrons, protons and neutrons in

atomd. Stability of atom

7. Two atoms are said to be isobars if:a. They have same atomic number but different mass

numberb. They have same number of electrons but different

number of neutronsc. They have same number of neutrons but different

number of electronsd. Sum of the number of protons and neutrons is same

but the number of protons is different

8. The number of radial nodes for 3p orbital is:a. 3 b. 4 c. 2 d. 1

9. Number of angular nodes for 4d orbital: a. 4 b. 3 c. 2 d. 1

10. Which of the following is responsible to rule out the existence of definite paths or trajectories of electrons?

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a. Pauli’s exclusion principleb. Heisenberg’s uncertainty principlec. Hund’s rule of maximum multiplicityd. Aufbau principle

11. Total number of orbitals associated with third shell will be:a. 2 b. 4 c. 9 d. 3

12. Orbital angular momentum depends on: a. l b. n and l c. n and m d. m and s

13. Chlorine exists in two isotopic forms. Cl-37 and Cl-35 but its atomic mass is 35.5. This indicates the ratio of Cl-37 and Cl-35 is approximately:a. 1 : 2 b. 1 : 1 c. 1 : 3 d. 3 : 1

14. The pair of ions having same electronic configuration

is: a. Cr3+, Fe3+ b. Fe3+, Mn2+c. Fe3+, Co3+ d. Sc3+, Cr3+

15. For the electrons of oxygen atom, which of the following statements is correct?a. Zeff for an electron in a 2s orbital is the same as Zeff for

an electron in a 2p orbitalb. An electron in the 2s orbital has the same energy as an

electron in the 2p orbitalc. Zeff for an electron in is orbital is the same as Zeff for an

electron in a 2s orbitald. The two electrons present in the 2s orbital have spin

quantum numbers ms but of opposite sign

16. If waves travelling at same speeds, which of the following matter waves have the shortest wavelength?a. Electron b. Alpha particle (He2+) c. Neutron d. Proton

41

Structure of Atom

1. Assertion: Element having 3d10 4s1 configurations are placed in s-block.

Reason: Last filling electron enters into s- sub shell.

2. Assertion: Orbital angular momentum of 1s, 2s, 3s are zero.

Reason: 1s, 2s, 3s all have spherical shape.

3. Assertion: If the potential difference applied to an electron is made 4 times, the De-Broglie wavelength associated is halved.

Reason: On making potential difference 4 times, velocity is doubled and hence λ is halved.

4. Assertion: Fe+2 has 24 electrons hence, its electronic configuration is similar to that of Cr(24)[Ar]3d5 4s1.

Reason: All the five unpaired electrons in 3d give stability to the ion.

5. Assertion: Electronic configuration of K(19) is 1s2,2s2, 2p6 3s2 3p6 4s1.

Reason: Energy of 4s < 3d hence, 4s is filled before 3d as decided by Aufbau principle.

6. Assertion: In an atom, the energy of the electron decreases as the value of n increases.

Reason: In an atom, the energy of 21en

− ∝ − .

7. Assertion: In third energy level there is no f subshell.

Reason: For n = 3, the possible value of are 0,1 and 2.

8. Assertion: Isotopes of an element have almost similar chemical properties.

Reason: Isotopes have same electronic configuration.

9. Assertion: The orbital angular momentum of 2s-electron

is equal to that of 3s-electron.

Reason: The orbital angular momentum is given by the relation ( )h 1

2+

π

and the value of it is same for

2s-electron and 3s-electron.

10. Assertion: The terms empirical formula and molecular formula bear the similar meaning.

Reason: Molecular formula of a compound defines the relative number of constituent atom in the simplest ratio.

11. Assertion: Boron always forms covalent bond.

Reason: The small size of B3+ favours formation of covalent bond.

12. Assertion: In H atom when electrons jump from 1s to 2s orbital, atom becomes cation.

Reason: H atom has only one electron.

13. Assertion: A spectral line will be observed for a 2px-2py transition.

Reason: The energy is released in the form of wave of light when electron drops from 2px to 2py orbital.

14. Assertion: The p-orbital is dumb-bell shaped.

Reason: Electrons present in p-orbital can have any one of the three values of magnetic quantum number i.e. +1, 0, –1.

15. Assertion: Energy of 3d-orbital is more than 4s-orbital for multi electron species.

Reason: Orbital energy is calculated by (n + 1) rule.

16. Assertion: Most of the α- particles striking gold foil in Rutherford’s experiment were deflected.

Reason: The size of nucleus is very small as compared

Assertion & Reason

Directions: These questions consist of two statements each, printed as Assertion and Reason. While answering these questions, you are required to choose any one of the following four responses.

A. If both Assertion and Reason are True and the Reason is a correct explanation of the Assertion.B. If both Assertion and Reason are True but Reason is not a correct explanation of the Assertion.C. If Assertion is True but the Reason is False.D. If both Assertion and Reason are False.

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to the size of atom.

17. Assertion: 2p-orbitals do not have any spherical node.

Reason: The number of spherical nodes in p-orbitals is given by (n-2), where n is principal quantum number.

18. Assertion: Wave number of a spectral line for an electronic transition is quantized.

Reason: It is proportional to the velocity of the electron undergoing the transition.

19. Assertion: Electronic energy for hydrogen atom of different orbitals follow the sequence:1s < 2s = 2p < 3s = 3p = 3d.

Reason: Electronic energy for hydrogen atom depends

only on n and is independent of ‘l’ & ‘m’ values.

20. Assertion: The radius of second orbit of He+ is equal to that of first orbit of hydrogen.

Reason: The radius of an orbit in hydrogen like species is directly proportional to n and inversely proportional to Z.

21. Assertion: In a multielectron atom, the electrons in different sub-shells have different energies.

Reason: Energy of an orbital depends upon n + l value.

22. Assertion: The number of angular nodes in 2z3d is zero.

Reason: Number of angular nodes of atomic orbitals is equal to value of l.

43

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Self Assessment Questions1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17d b b b b b b d a a a b c d d c c18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34a a b c d c a b d b c a c d a b a

35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51d b a d d d c d d b b b c a d b a52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68a a b c d a b a a d d c d c a a c

69 70 71 72 73 74 75a b a b c c c

Higher Order Questions

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17d b d a c d b d d c d d d d a c c18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34a b a d a b c a c b a c d d c b c

35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51c d a c b c c b c d a d a c c c d

52 53 54d d c

NCERT Exemplar Problems

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16c b d d b a d d c b c a c b d b

Assertion & Reason

1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17d b a d a d a a a c a d d b a d a18 19 20 21 22c a d a b

Answer Key

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Self Assessment Questions

1. (d) It was determined by Millikan's with a help of oil drop experiment.

2. (b) J.J. Thomson was the first one to demostrate particle nature of electron.

3. (b) Because they both have 2 electrons

4. (b) Number of neutrons = M – ANumber of neutrons in 14Si

30 = 30 – 14 = 16 Number of neutrons in 16Si

32 = 32 – 16 = 16Because of the presence of same number of neutrons, 14Si

30 and 16S32 are called isotones.

Hence; 32Ge 76 , 34Se

76 and 14Si30 and 16S

32 are the isobars or isotones.

5. (b) Positron is lightest. 6. (b) The ratio of charge and mass would be greater for

electron as the mass of electron is least among the species.

7. (b) Neil Bohr utilised the concept of quantisation of energy (proposed by max planck) first time to give a new model of an atom.

8. (d) ( )( ) ( )

2H 2 2

1 2

1 1 1R Zn n

= − λ

where, RH = R, Z = 3 ( for Li2+, atomic no. 3)

n2 = 2, n1 = 1

( )( ) ( )

22 2

1 1 1R 31 2

= − λ

R 9 3 4

4 27R× ×

= ⇒ λ =

9. (a) Zeeman effect is the splitting of spectral line emitted by an excited electron under the influence of magnetic field.

10. (a) Weight of 200X = 0.90 × 200 = 180.00 u

Weight of 199X = 0.08 × 199 = 15.92 uweight of 202X = 0.02 × 202 = 4.04 uTotal weight = 199.96u ≈ 200 u

11. (a) 109,677 cm–1. 12. (b) Isoelectronic species have same number of electrons.

19K+: Number of electrons = 19 – 1 = 18

17Cl– : Number of electrons = 17 + 1 = 18

20Ca2+: Number of electrons = 20 – 2 = 18

hence, K+, Cl– and Ca2+ are isoelectronic species.

13. (c) We know that H 2 21 2

1 1E hc Rn n

∆ = −

For the lowest energy of spectral line in the lyman series.n1 = 1, n2 = 2Hence,

( ) ( )H 2 21 1E hc R1 2

∆ = −

H1 1hcR1 4 = −

H3hcR4

=

14. (d) In 1885, Balmer for the first time showed that the wave numbers of spectral lines present in the visible region in hydrogen spectrum are given by:

( )( )

12 2

2

1 1cm 109677n2

− ν = −

Hence; n2 = 3, 4, 5 ........Hence; the Balmer spectrum of hydrogen was discovered first and it lies in visible region.

15. (d) According to Bohr, velocity of an electron (v) 6 Z2.188 10 m/ sec

n= ×

for H ; Z = 1 n = 2 (given)6

62.188 10 1V 1.094 10 m/sec2× ×

∴ = = ×

16. (c) Bohr model of an atom is applicable for those species which have only 1 electron but He2+ do not have any electron. Therefore, this model is not applicable to He2+

17. (c) Ionisation energy of an atom of atomic number, Z is given by

( ) 2 2H2 2

I.E Z ZI.E. 13.6 eVn n×

= = ×

Thus, second ionisation energy for He-atom is given by

Explanations

45

Structure of Atom

( )( )

2

2

13.6 2I.E. eV 54.4 eV

1×

= ⇒

18. (a) The lines are called Balmer series. 19. (a) Nitride ion can be represented as N3–.

Number of protons = 7Number of electrons = 7 + 3 = 10Therefore, nitride ions contain 7-protons + 10 electrons.

20. (b) 39K+ and 40K+ both contain 18 electrons; i.e., they are isoelectronic and isotopic as they have same atomic number.

21. (c) Neutrons in 126 C 6=

Neutrons in 2814 Si 14=Ratio : 6 : 14 = 3 : 7

22. (d) 16 p means Z = 16, 18e– means 2 unit negative charge is present.Hence; species is S2–.

23. (c) When an electron jumps from higher energy level n = 4 to ground state (i.e., n = 1)

The number of spectral lines( )( )n n 1

2−

=

∴ When an electrons jump from n = 4 level, number of

spectral line: ( )( )4 4 1 4 3 12 62 2 2− ×

= ⇒ = =

24. (a) Wave number of spectral lines in emission series of hydrogen,

H 2 21 2

1 1Rn n

ν = −

Given; ( )H8 R9

ν =

On putting the value of ν in equation (i), we get

H H 2 21 2

8 1 1R R9 n n

= −

( )2 228 1 19 n1= −

2 22 2

8 1 1 119 n 9 n

− = ⇒ =

n2 = 3 ∴ electrons jumps from n2 = 3 to n1 = 1

25. (b) Radius of H-like species,

As 2

n0.529 nr

Z×

= is radius for first Bohr orbit of H-atom

which is equal to r (given).

So, radius of first orbit of Li2+.(n = 1, Z = 3, for Li2+) is r/3

26. (d) In OH–, number of protons = 8 + 1 = 9number of neutrons = 8 + 0 = 8number of electrons = 8 + 1 + 1 = 10Hence, in OH– the number of protons is greater than the number of neutrons but number of protons is less than the number of electrons.

27. (b) o o

1 23000A, 6000Aλ = λ =

1 21 2

hc hc hc hcE , E3000 6000

= = = =λ λ

1

2

E hc 6000 2 :1E 3000 hc

= × =

28. (c) As the Bohr's model applicable only to H-atom and H-like species containing one electron system.

29. (a) Rydberg ritz equation;2 21 2

1 1Rn n

ν = −

for Balmer series; n1 = 2, n2 = 3

19 4 5RR cm36 36

−− ν = =

30. (c) Second Bohr orbit of hydrogen atom, i.e., n = 2.Atomic number of hydrogen (z) = 1

by using ( )22 0.529 20.529nr

z 1×

= =

o

2.116 A 0.2116 nm= ⇒

31. (d) As En 2

192

z21.8 10n

−= − × ×

Thus; n 21E

n− ∝

where n is the number of orbit. Hence;

as the value of "n" increases, energy of the electron also increases; due to this reason; maximum energy is possessed by an electron when it is present at infinite distance from the nucleus.

32. (a) hcE =λ

1E∴ ∝λ

∴ Decrease in wavelength, increases the energy.

Energy difference in n4 → n1 transition is maximum.

∴ n4 → n1, transition has minimum wavelength

33. (b) 2

3 2 12 1

r H r H 2r Be r H4 4

+ ×= = =

34. (a)

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35. (d) 36. (b)

37. (a) Total

KE 1E

= −

Total energy = –3.4 eV (Given)

∴ KE = –(–3.4 eV) = +3.4 eV 38. (d) 20Ca = 1s2 2s2 2p6 3s2 3p64s2

20Ca2+ = 1s2 2s2 2p6 3s2 3p6

Hence, Ca2+ has 8 electrons in outermost as well as in penultimate shell.

39. (d) According to Heisenberg's uncertainty principlehm v x

4∆ ×∆ ≥

π

m = 6.626 × 10–28 kg∆v = 10–6 m/sech = 6.626 × 10–34 Js

3428 6 6.626 106.626 10 10 x

4

−− − ×× × ×∆ =

π34

34

6.626 10 1x44 6.626 10

−

−

×∆ = =

ππ× ×

40. (d) hmv

λ =

mass of α–particle = 4u, mass of deuterium = 2uVD = 5Vα

DD

1 1,2 V 4Vα α

λ = λ =×

D

D

4V 4 4 0.42V 2 5 10

α

α

λ= = = =

λ ×

41. (c) According to de-Broglie equation;Wavelength ( ) h

mvλ =

Given; m = 100g, v = 100 cm/sec

h = 6.6 × 10–34 Js = 6.6 × 10–27 erg-s

on substituting values, we get27 27

314

6.6 10 6.6 10 6.6 10 cm/s100 100 10

− −−× ×λ = = = ×

× 42. (d) According to de-Broglie equation

34 3429

6 5h 6.63 10 6.63 10 6.63 10 m

mc 10 10 10

− −−

− −

× ×λ = = ⇒ = ×

×

43. (d) de-Broglie equationh hv .....(1)

mv mλ = ⇒ =

λ

K.E. = 1/2 mv2 .....(2)

on putting the value of v in e. (2) we get2 2

2

1 h 1 hK.E. m2 m 2 m

= = λ λ

Hence, K.E. ∝ 1/m (if λ is same) and order of mass is asme < mp < mαThus; the order of K.E. isEe > Ep > Eα

44. (b) According to Sommerfeld's modification, the trajectory of an electron in a hydrogen atom is a closed ellipse like curve in addition to circular orbits.

45. (b) Heisenberg gave the uncertainty principle in 1920. According to Heisenberg, it is impossible to determine preciously both the position and momentum or velocity of a small moving particle, eg: proton and electron.

46. (b) 27 27

293 5 2

h 6.626 10 6.626 10 6.626 10 cmmv 10 10 10

− −−

−

× ×λ = = ⇒ = ×

×

47. (c) According to Heisenberg principlehx

4 m∆ =

π ∆ν27

28 4

6.6 10 100x4 3.14 9.1 10 3 10 0.011

−

−

× ×∆ =

× × × × × ×

∆x = 0.175 cm

48. (a) Here; ∆x = ∆vhx p

4∆ ×∆ ≥

πhx m v

4∆ × ∆ =

π2 h( v)

4 m∆ =

π

(∴ ∆x = ∆v)

hv4 m

∆ =π

∆p = m. ∆v

h mhp m4 m 4

∆ = =π π

1 mhp2

∆ =π

49. (d) hmv

λ =

A BA A

h h,m 0.1 5 m 0.05

λ = λ =× × ×

A A

B A

m 0.05 5 5m 0.1 2

λ × ×= =

λ ×

47

Structure of Atom

50. (b) Let, the wavelength of particle be x, Thus; x = 100 × velocity

velocity (v) x h,100 mv

= λ =

h 100xm x×

=×

2 h hx 100 x 10m m

= × ⇒ =

51. (a) de-broglie equation is hmv

λ =

2 2K.E.1K.E. mv or v2 m= =

h2m K.E.

λ =

221 2

22 1

K.E 5 25K.E 3 9

λ = = = λ

52. (a) According to Heisenberg's uncertainty principle hx p

4∆ ×∆ ≥

π

Here ∆x = ∆p and ∆p = m.∆v

22

h 1 hv or v2mm 4

∴ ∆ = ∆ =ππ

53. (a) 34h 6.626 10

2mE 2 1 0.5

−×λ = =

× ×

346.626 10 m.−= ×

54. (b) Wave nature of electrons was first demonstrated by de-Broglie who gave the following equation for the wavelength of electrons.

hmv

λ =

55. (c) According to Heisenberg's uncertainty principlehx. p

4∆ ∆ ≥

π

But ∆p (momentum) = m∆vhx.m v

4∆ ∆ ≥

πhx

4 m v∆ =

π ∆

56. (d) P(1s2, 2s22p6, 3s2 3p3) has 6 electrons in s-subshells as in d-shell of Fe2+.

57. (a) If m = + 3 (maximum), then = 3 and n can have maximum value of four. Also no. of waves in an orbit = no. of orbit.

58. (b) for any value of the possible value of m = – to + = 1, m = –1, 0, +1

59. (a) The quantum of light energy is called photon.

60. (a) K.E. in an orbit 2

0

Ze .....(i)8 r

=πε

Magnitude of P.E. in an orbit 2

0

Ze .....(ii)4 r

=πε

On comparing (i) & (ii); we get

1K.E. P.E.2

=

61. (d) Orbital angular momentum ( )h 12

= +π

For f-orbital, = 3∴ orbital angular momentum ( )h 3 3 1

2= +

π

( )h 3 42

=π

2h 3 h 32

= =π π

3h=

π

62. (d) Electronic configuration ofNi-28 is [Ar] 3d8, 4s2

Thus, outer orbital diagram for Ni2+ is

⇒ n = 2

n(n 2) B.M.

2(2 2) B.M.

8

µ = +

µ = +

µ =

µ = 2.83 B.M. 63. (c) The maximum number of electrons which can be

held by electron with Azimuthal quantum number "" in an atom is given by 2(2 + 1).

64. (d) The orbital of the electron having n = 3, = 1, m = –1 is 3px (as n, , m) and an orbital can have a maximum of two electrons with opposite spins.m = –1 (3px), 0 (3py) and +1(3pz),Therefore, 3px orbital contain only two electrons.

65. (c) Azimuthal quantum number determines shape of the orbital magnetic quantum number, m is associated with spatial orientation of the orbital. Principal quantum number (n) determines energy of the electron & size (volume) of the atom.

66. (a) Atomic number of element having electronic configuration 2 2 1 1 1x y z1s 2s 2p 2p 2p is 7. The atomic number 7 is of nitrogen atom.

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67. (a) Atomic number 17, 1s22s22p63s23p5

Valence shell has 3s2 3p5 orbit.

Valence shell has 3 orbitals with paired electrons.

68. (c) Electronic configuration of bromine is

1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p5

Then; the total no. of p–electron in bromine is 17.

69. (a) Electronic configuration of 26Fe is

1s2 2s2 2p6 3s2 3p6 3d6 4s2

Fe3+ = 1s2 2s2 2p6 3s2 3p6 3d5

⸪ Outer orbital diagram of Fe3+ is

∴ the no. unpaired electrons are 5.

70. (b) Orbital angular momentum ( ) h12

= +π

⸪ for p-electron = 1

orbital angular momentum ( ) h1 1 12

= +π

2h h2 2

= =π π

71. (a) The hydrogen atom has configuration 1s1.

So; 3s, 3p & 3d orbital will have the same energy with respect to 1s orbital.

72. (b) nhL2

=π

n = 44h 2hL2

∴ = =π π

73. (c) Total no. of orbital in a shell is n2.

74. (c) Radius of Bohr's orbit2

0n az×

=

20

0(3) a

9a1×

= =

75. (c) According to azimuthal quantum number

= n – 1

= 2 – 1 = 1

But in option (c), = 0

so that it is incorrect.

Higher Order Questions

1. (d) NCERT (XI) Ch - 2, Pg - 442nd excited state refers to 3rd shell

∴ 2

2

13.6 Z 243×

=

2 24 9Z

13.6×

=

Now, 2

2

ZIE 13.6n

= + ×

2

24 9IE 13.613.61

×= ×

24 9IE 13.6 216eV13.6×

= × =

2. (b) NCERT (XI) Ch - 2, Pg - 462πr = nλFor minimum radius, n = 12πrmin = λ

minr 2λ

=π

3. (d) NCERT (XI) Ch - 2, Pg - 37.Formula for energy of photon is already mentioned, i.e.

= o

o

12400eV A

(A)λo

o

12400eV AE600 10(A)

=×

(1nm=10Å)

12400hv eV6000

=

19124hv 1.6 10 J60

−= × × (1 eV = 1.6 × 10–19 J)

34 191246.626 10 Js 1.6 10 J60

− −× ×ν = × ×

v = 5 × 1014 s–1

To calculate energy of a photon,o

o

12400eV AE 2.07eV6000A

= =

4. (a) NCERT (XI) Ch - 2, Pg - 37E = hv

34 8

9

6.626 10 3 10E J310 10

−

−

× × ×=

×

49

Structure of Atom

E = 0.0641 × 10–26 × 109 JE = 0.0641 × 10–17 JThis is the energy of one photon required to break one A2 molecule.To break 1 mole of A2, Energy of photons required =

23 17 66.02 10 0.0641 10 0.386 10 386 kJ/mole−× × × = × =

K.E = 386 kJ/mole – 288 kJ/mole = 98 kJ/molePercentage of total energy that got converted into KE = 98 100 25%

386× =

5. (c) NCERT (XI) Ch - 2, Pg - 442

nnr 0.529z

= ×

1

2

21

n2

n 2

n0.529r4 z9 r n0.529

z

×= =

×

2

1

2

n49 n

=

1

2

n23 n=

n1 could be = 2n2 could be = 3 orn1 could be = 4

n2 could be = 6Energy difference between 3rd and 2nd shell =

2 2

2 2

1 113.6 13.63 2

− × − − ×

13.6 13.6 1 113.69 4 4 9

513.6 1.89eV36

− = + = −

= × =

Energy difference between 6th & 4th shell = 2 2

2 2

1 113.6 13.66 4

= − − −

13.6 13.6 1 113.636 16 16 36

− = + = −

= 0.49 eV 6. (d) NCERT (XI) Ch - 2, Pg - 36

(Fact Based) 7. (b) NCERT (XI) Ch - 2, Pg - 42

Last line of lyman series refers to transition ∞ → 1

2 21

1 1 1R1 = − λ ∞

1

1 R=λ

......(1)

2nd line of Balmer series refers to transition 4 → 2.

2 22

1 1 1R2 4

= − λ

2

1 1 1R4 16

= − λ

2

1 3R16

= ×λ

......(2)

1

2

1R R 16

1 3R 1 3R16

λ= = ×

λ

2

1

163

λ= =λ

8. (d) NCERT (XI) Ch - 2, Pg - 51 & 52n = 4l = 2 (d-sub shell)m = 1 (- to +)s = –1/2

9. (d) NCERT (XI) Ch - 2, Pg - 51 = 2 refers to d-subshell(4d is possible) = 3 refers to f-subshell(4f is also possible)

10. (c) NCERT (XI) Ch - 2, Pg - 43 & 423rd line of Balmer seriesrefers to transition 5 → 2

Change in angular momentum = 5h 2h2 2

−π π

3h2

=π

11. (d) NCERT (XI) Ch - 2, Pg - 46Let initial state be n1

1n

h2m K.E

λ =

1

1n

h2m K.E

λ =

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1

22

1n

h2m K.E

λ =

1

2

n 21

hK.E.2m

=λ

1

2

n 21

hT.E.2m−

=λ

------ (1)

T.E in n2 (where λ = λ2)

2

2n

h2mKE

λ =

2

22

2n

h2mKE

λ =

2

2

n 22

hK.E2m

=λ

2

2

n 22

hT.E2m−

=λ

-------(2)

1 2n nhcT.E T.E− =λ

2 2

2 21 2

h h hc2m 2m

− −− = λλ λ

2

2 22 1

h 1 1 hc2m

− = λλ λ

2 221 2

2 21 2

h hc2m

λ − λ= λλ λ

2 21 2

2 21 2

2mch

λ λλ = λ − λ

12. (d)

The transition 3 → 2 represents first line of Balmer series Now, IE of H = 13.6 eVIt’s given as x J/atomSo, T.E. of an e– =

2

2zx J / atomn

− ×

2 2

3 2 2 2

2 2E E E x x3 2

∆ = − = − × − − ×

E x x94

∆ = − × +

4 9x 4xE x x9 9

−∆ = − = ⇒ 5xE

9∆ =

∴ Energy of photon emitted = 5x9

Apply hc 5x hcE9

= ⇒ =λ λ

9hc5x

λ =

13. (d) He → He+ + e–; IE1 = 24.6 eVHe+ → He2+ + e–; IE2 = ?He+ is an unielectron specie for which we can apply,

2

2

zIE 13.6 eV / atomn

= + ×

2

2

2IE 13.6 54.4eV1

= + × =

So, total energy required to remove both the electrons = 24.6 + 54.4 = 79 eV

14. (d)

Apply, hν – hνo = kE

Apply, o3h h h4

ν − ν = ν ⇒ o3h h h4

ν − ν = ν

oh h4ν= ν ⇒ o4

ν= ν

1615

o3.2 10 8 10 Hz

4×

= ν = ×

15. (a) 2

2

zT.E. Rhcn

= − ×

2

Rhc 4Rhc4 n

−= − ×

n2 = 16 ⇒ n = 4 16. (c) H–(2):- 1s2

It's a multielectron specie, first excited state will be 2s & 2nd excited state will be 2p. Degeneracy of 2s is 1. Degeneracy of 2p will be 3.

17. (c)

18. (a) ( ) hcE in eV∆ =λ

.34 8 119 6.626 10 Js 3 10 msE 1.6 10 J

− −− × × ×∆ × × =

λ

51

Structure of Atom

34 8

19

6.626 10 3 10 m1.6 10 F

−

−

× × ×λ =

× ×∆o

34 8 10

19

6.626 10 3 10 10 A1.6 10 E

−

−

× × × ×λ =

× ×∆o

34 18 1912.42 10 10 10 AE

−× ×× ×λ =

∆o

34 3712.42 10 10 AE

−× ×λ =

∆

o3 o o12.42 10 A 12420 12375A A

E E E×

λ = = ≈∆ ∆ ∆

19. (b) 2

2

ZT.E 13.6 eV / atmn

= − ×

2

2

1T 3.4 13.6n

= − ×

n2 = 4n = 2 (first excited state)

20. (a) 8 1n

zV 2.18 10 m sn

−= × ×

or6 1

nzV 2.18 10 msn

−= × ×

1n o

zV V msn

−= ×

8 6 1o

1V 3 10 3 10 ms100

−= × × = ×

6 1n

zV 3 10 msn

−= × ×

Velocity in 5th Bohr orbit

th6 1 6 1

5

1V 3 10 ms 0.6 10 ms5

− −= × × = ×

Velocity relative to speed of light 6

2 38

0.6 10 0.2 10 2 10 0.0023 10

− −× = × = × =×

21. (d) Let formula for I.E. = 2

2

Zx J / atomn

+ ×

19.6 × 10–18 = 22x

12+ ×

19.6 × 10–18 = x × 41819.6 10 x

4

−×=

x = 4.9 × 10–182

182

z4.9 10 J/ atomn

−× ×

218

2

z4.9 10 J / atomn

−− × ×

(⸪ T.E. & I.E. have opposite signs & rest formula is the same)

218 18

2

34.9 10 4.9 10 91

− −− × × = − × ×

= –44.1 × 10–18 J atom–1

22. (a) hν1 – hνo = kE1 .......(1)hν2 – hνo = kE2 ........(2)

( )( )

1 0

2 0

h 2h 1ν − ν

=ν − ν

ν1 – ν0 = 2ν2 – 2ν0 ⇒ ν0 = 2ν2 – ν1

23. (b) nd nd nd

st st st

2 22 orbit 2 orbit 2 orbit

2 2I orbit 1 orbit I orbit

Area r rArea r r

π= =π

22

2

20.529z 16

1120.529z

×

= = ×

24. (c) nhmvr .......(1)2

=π

hp mv ......(2)= =λ

Placing the value of mv from Eq. (2) into (1) for 3rd orbit

3h 3hr

2=

λ π3

23 o o

2 r r n r 9r3π

λ = ⇒ = =

oo

2 .9rSo, 9 r

3π

λ = = π

25. (a) PE = –2KE

2xPE will change from 2x to4

∴ − −

( )2xChange in potential energy 2x4

= − − −

x 3x2x2 2

= − + =

26. (c) H 2 21 1 1R

2 n = − λ

for first line n = 3,

2 2

1 1 11096782 3 ∴ = − λ

λ = 6569 Å.

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27. (b) 1H 1H 1H3 2E E 5EE E for H9 4 36

− = + =

1Hn 2

EEn

=

2 21H 1H

6 4 2 2

E 2 E 2E E for He6 4

+ × ×− = − +

21H 1H

2nHe

E z 5EE36n+

×= = +

28. (a) 2 21 1 22 1 1

E E3, E :E 25 : 9E 5 E

λ= = ∴ =

λ

29. (c) For numerical purposes,hx. p

4∆ ∆ =

π ⇒

hx. m v4

∆ ∆ =π

hx. v4 m

∆ ∆ =π

⇒ hx.4 m

∆ ∆ν =π

2 10.005 600 3 10 ms100

− −∆ν = × = ×

34

31 2

h 6.626 10x4 m 4 3.14 9.1 10 3 10

−

− −

×∆ = =

π ∆ν × × × × ×Δx = 1.92 × 10–3 m

30. (d) Energy absorbed - Bond ENERGY = k.E of the molecule4.4 × 10–19J – 4 × 10–19J = kE of molecule10–19 × 0.4 J = kE of molecule4 × 10–20 J = kE of molecule

k.E. per atom = 20

204 10 2 10 J2

−−× = ×

31. (d)

32. (c) 1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6

33. (b) For IInd periodn = 2 henceI m s

0 0 1 1, 0,2 2

+ −

1 –1 1 1, 0,2 2

+ −

0 1 1, 0,2 2

+ −

+1 1 1, 0,2 2

+ −

Hence, total number of elements (total value of spin quantum number) = 12

34. (c) Violet line ⇒ λmin ⇒ Emax = 4 → 1 35. (c) Option 1 not possible because s can never have –2

valueOption 2 not possible because n and l cannot have same valueOption 4 not possible because s cannot have zero value∴ Correct answer = 3

1n 4, l 3, m 3,s2

= = = − =

36. (d) Total number of e– = 30. Therefore, e– configuration will be

1s2 2s2 2p6 3s2 3p6 4s2 3d10

n = 1 n = 2 n = 2 n = 3 n = 3 n = 4 n = 3l = 0 l = 0 l = 1 l = 0 l = 1 l = 0 l = 2m = 0 m = 0 m = –1,

0, +1m = 0 m = 0,

–1, +1m = 0 m = –2,

–1, 0, +1, +2

For s-subshell 1 orbital have m = 0For p-subshell 1 orbital have m = 0For d-subshell 1 orbital have m = 0

∴ Total 7 orbital have m = 0 in above configuration. Therefore, total number of electron = 7 × 2 = 14

37. (a) Correct graph will be

38. (c) m = represents the orientation of orbital in magnetic field.m = orbitals

39. (b) (ii), (iv), and (v) are not possible

(ii) n = 2 l = 2 m = 1 s = +1/2 l not equal to n not possible

(iii) n = 1 l = 0 m = –1 s = –1/2 Not possible because m = –1 where l = 0

(iv) n = 3 l = 2 m = 3 s = +1/2 Not possible because m = 3 is not for l = 2

40. (c) Any orbital have maximum of two electrons with opposite spin.

53

Structure of Atom

41. (c) When n = 3, cannot be 3, so (a) is incorrect when l = 1, m cannot be = +2.

42. (b) n2 2

313.6 313.6E kcal / mol 78.4n n

n 2

− −= ⇒ − =

∴ =

43. (c) 2 2

2 2

n hr4 mZe

=π2

23 22

3

r 2 9r rr 43

∴ = ∴ =

44. (d) rn ∝ n2

But rn + 1 – rn = rn – 1(n + 1)2 – n2 = (n – 1)2

n = 4

45. (a) 3

23

hc 430.53 10 K.E.6.023 10

×= +

λ ×34 8 3

209 23

6.626 10 3 10 430.53 10K.E. 6.9 10253.7 10 6.023 10

−−

−

× × × ×= − =

× ×

20

19

6.9 10Fraction 0.088 8.86%7.83 10

−

−

×∴ = = =

×

46. (d) When intensity is doubled, number of electrons emitted per second is also doubled but average energy of photoelectrons emitted remains the same.

47. (a) (b) is wrong because shape is given by azimuthal quantum number and magnetic quantum number tells the orientation. (d) is wrong because electrons in different shells travel with different velocities.

48. (c) In case of hydrogen like atoms, energy depends on the principal quantum number only. Hence, 2s-orbital will have energy equal to 2p-orbital.

49. (c) 7632 Ge : n = 76 – 32 = 44

i. 7732 Ge : n = 77 – 32 = 45

ii. 7733 As : n = 77 – 33 = 44

iii. 7734 Se : n = 77 – 34 = 43

iv. 7834Se : n = 78 – 34 = 44

50. (c) The orbitals described by these sets of quantum numbers are :

(i) 4s (ii) 3p (iii) 3d (iv) 3s

The energy of these orbitals follows the order :

3d > 4s > 3p > 3s

(iii) (i) (ii) (iv)

51. (d) a. Co3+ : 1s22s2 2p63s23p63d6 ∴ 4 unpaired electrons

( )4 4 2 24 4.9BM∴ µ = + = = b. Number of radial = n – – 1 Number of radial nodes in 3p orbital = 3 – 1 – 1 = 1c. Number of electrons with (m = 0) in Mn2+ (1s22s22p63s23p63d5) ion = 1 s (2) + 2s(2) + 2p(2) + 3s(2) + 3p(2) + 3d(1) = 11d. Orbital angular momentum for the unpaired electron in V4+ lies in 3d orbital. ∴ = 2

∴ Orbital angular momentum = ( ) h 6h12 2

+ =π π

NCERT Exemplar Problems

1. (c) Bohr put forward concept of electrons move in a circular path of fixed energy called orbits but not derived from Rutherford’s scattering experiment.

2. (b) Correct configuration should be 1s2 2s2 2p6 3s23p6 3d10 4s1 for the copper which has atomic number 29 (29Cu). Due to extra stability of full filled orbital of d-subshell, the last electron enter into d-orbital instead of s-orbital.

3. (d) The probability density of electrons in 2s orbital first increases then decreases an after that it begins to increases again as distance increases from nucleus.

4. (d) Negatively charged material particles are present in cathode rays

5. (b) The mass of electron is very small as compared to the mass of the neutron.

Mass of electron = 9.1 × 10–31 kg

Mass of neutron = 1.67 × 10–27 kg

6. (a) J J Thomson, in 1898, proposed plum pudding, (raisin pudding or watermelon) model of atom. In this model mass of atom is assumed to be uniformly distributed. This model explain the overall neutrality of the atom.

7. (d) Isobars have the same mass number (i.e., sum of protons and neutrons) but different atomic number (i.e., number of protons). e.g., 18Ar

40 and 19K40 are isobars.

18Ar40

19K40

8. (d) For a hydrogen atom wave function, there are n – – 1 radial nodes and (n – 1) total nodes

Number of radial nodes 3p orbital

= n – l – 1

= 3 – 1 – 1 = 1

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9. (c) Number of angular nodes =

For 4th orbital (n = 4) and = 2 for d-orbital

Number of angular nodes = 2

10. (b) A German physicist Werner Heisenberg in 1927, stated uncertainty principle which states that it is not possible to determine simultaneously, the exact position and exact momentum of an electron.

Mathematically, h4

∆ ×∆ ≥π

x

The important implications of the Heisenberg uncertainty principle is that it rules out existence of definite paths or trajectories of electrons and other similar particles.

11. (c) Total number of orbitals associated with nth shell = n2

∴ Total number of orbitals associated with third shell (3)2 = 9

12. (a) Orbital angular momentum

( )2 1=

+

hmvrl l

Hence, ‘l’. l can have values ranging from 0 to (n – 1) it depends only on.

(a) When l = 0, the subshell is s and orbital is spherical in shape.

(b) When l = 1, the subshell is p and orbital is dumb-bell shaped.

(c) When l = 2 , the subshell is d and orbital is double dumb-bell shaped.

(d) When l = 3, the subshell is f and orbital is complicated in shape.

13. (c) The fractional atomic mass (35.5) of chlorine is due to the fact that in ordinary chlorine atom. Cl-37 and Cl-35 are present in the ratio of 1 : 3.

∴ Average atomic mass of 3 35 1 37 105 32Cl 35.5 amu

4 4× + × +

= = =

14. (b) 24Cr = [Ar]3d5, 4s1 24Cr

3+ = [Ar]3d3

26Fe = [Ar]3d6, 4s2 26Fe

3+ = [Ar]3d5

25Mn = [Ar]3d5, 4s2 25Mn

2+ = [Ar]3d5

27Co = [Ar]3d7, 4s2 27Co

3+ = [Ar]3d6

21Sc = [Ar]3d1, 4s2 21Sc

3+ = [Ar]

Thus, Fe3+ and Mn2+ have the same electronic configuration

15. (d) (a) Electrons in 2s and 2p orbitals have different screen effect. Hence, their Zeff is different. Zeff of 2s orbital > Zeff of 2p orbital.

Therefore, it is not correct.

(b) Energy of 2s orbital < energy of 2p orbital.

Hence, it is not correct.

(c) Zeff of 1s orbital ≠ Zeff of 2s orbital

Hence, it is incorrect.

(d) For the two electrons of 2s orbital, the value of ms is 1 1and .2 2

+ −

16. (b) From De-Broglie equation,wavelength, λ = hmv

For same speed of different particles, i.e., electron, proton, neutron and α-particle,

1λ ∝

m

As h is constant. Greater the mass of matter waves, lesser is wavelength and vice-versa. In these matter waves, alpha particle (He2+) has higher mass, therefore, shortest wavelength.

Assertion & Reason

1. (d) [Ar]3d104s1 is configuration of Cu ([Ar]3d9 4s2 → ([Ar]3d10 4s1

2. (b) In the ground state there is always spherical symmetry in an atom

3. (a) Let Ve = Voltage v = electron velocity

2e

e2VmvV V

2 m= ⇒ =

4. (d) Fe2+ → half filled and full filled configuration is more stable than rest other configurations

3d 4S 4P 3d 4S

5. (a) Both Assertion and Reason are True and the Reason is a correct explanation of the Assertion

6. (d) As energy is required in shifting away every negative charged electron from +ve charged nucleus, the energy of orbital will increase with increase in n

7. (a) The max no. of electrons in a energy level (n) is equal to 2n2. For n = 3, no. of e = 18 which is equal to the

55

Structure of Atom

total no. of electron in the s, p & d orbitals

8. (a) Isotopes are elements having same atomic no. i.e.some no. of electrons in the atom, hence same electronicconfiguration

9. (a) Since all a orbitals have same L value i.e O,therefore all a orbitals have O angular momentum oftheir electrons

10. (c) Empirical formula represents simplest whole no ratioof various atoms present in compound

11. (a) Due to its small size, sum of 1st 3 ionization enthalpies is very high. This prevents it to form +3 ions and forcesit to form doubly covalent compound.

12. (d) The atom gets into 1st excited state as electronmakes transition from 1st energy level to 2nd energylevel

13. (d) Since they are degenerate orbitals no energy will bereleased in this transition

14. (b) Each p-orbital consist + of 2 sections (lobes) oneithes side of plane that pass through nucleus

15. (a) By n + l rule, since d orbital has l = 2 & for s, l = 0

∴ (n + l)3d > (n + l)4s hence A is true

16. (d) Most of the α particles passed through gold foilundeflected

17. (a) In 2p orbital there is only inode (planar) separating 2lobes of a p orbital.

18. (c) Wave no of a spectral line depends upon initial andfinal state of the electron making transition which isfixed for a particular spectral line

19. (a) Both Assertion and Reason are True and the Reasonis a correct explanation of the Assertion.

20. (d) Both Assertion and Reason are False.

21. (a) Both Assertion and Reason are True and the Reasonis a correct explanation of the Assertion.

22. (b) Both Assertion and Reason are True but Reason isnot a correct explanation of the Assertion.

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